In the event that you are reading such a version, I have a request:

Kinematics, Part 1

Special Relativity, For the Enthusiastic Beginner (Draft version, December 2016)

Copyright 2016, David Morin, morin@physics.harvard.edu

TO THE READER: This book is available as both a paperback and an eBook. I have made the first chapter available on the web, but it is possible (based on past experience) that a pirated version of the complete book will eventually appear on file-sharing sites.

In the event that you are reading such a version, I have a request:

If you don’t find this book useful (in which case you probably would have returned it, if you had bought it), or if you do find it useful but aren’t able to afford it, then no worries;

carry on. However, if you do find it useful and are able to afford the Kindle eBook (priced below $10), then please consider purchasing it (available on Amazon). If you don’t already have the Kindle reading app for your computer, you can download it free from Amazon. I chose to self-publish this book so that I could keep the cost low. The resulting eBook price of around $10, which is very inexpensive for a 250-page physics book, is less than a movie and a bag of popcorn, with the added bonus that the book lasts for more than two hours and has zero calories (if used properly!).

– David Morin

Special relativity is an extremely counterintuitive subject, and in this chapter we will see how its bizarre features come about. We will build up the theory from scratch, starting with the postulates of relativity, of which there are only two. We will be able to derive a surprisingly large number of strange effects from these two easily stated postulates.

The postulate that most people find highly counterintuitive is that the speed of light has the same value in any inertial (that is, non-accelerating) reference frame. This speed, which is about 3 · 10

⁸

m /s, is much greater than the speed of everyday objects, so most of the consequences of relativity aren’t noticeable. If we instead lived in a world identical to ours except for the speed of light being only 50 mph, then the consequences of relativity would be ubiquitous. We wouldn’t think twice about time dilation, length contraction, and so on.

As mentioned in the preface, this chapter is the first of two that cover kinematics.

(Kinematics deals with lengths, times, speeds, etc., whereas dynamics deals with masses, forces, energy, momentum, etc.) The outline of this chapter is as follows. In Section 1.1 we discuss the historical motivations that led Einstein to his theory of special relativity.

Section 1.2 covers the two postulates of relativity, from which everything in the theory can be obtained. Section 1.3 is the heart of the chapter, where we derive the three main consequences of the postulates (loss of simultaneity, time dilation, and length contraction). In Section 1.4 we present four instructive examples that utilize the three

1

fundamental effects. Section 1.5 covers the velocity-addition formula, which gives the proper correction to the naive Newtonian result (simply adding the velocities). In Chapter 2 we will continue our discussion of kinematics, covering more advanced topics.

1.1 Motivation

Although it was certainly a stroke of genius that led Einstein to his theory of relativity, it didn’t just come out of the blue. A number of conundrums in 19th-century physics suggested that something was amiss. Many people had made efforts to explain away these conundrums, and at least a few steps had been taken toward the correct theory.

But Einstein was the one who finally put everything together, and he did so in a way that had consequences far beyond the realm of the specific issues that people were trying to understand. Indeed, his theory turned our idea of space and time on its head. But before we get to the heart of the theory, let’s look at two of the major problems in late 19th-century physics. (A third issue, involving the addition of velocities, is presented in Problem 1.15.) If you can’t wait to get to the postulates (and subsequently the results) of special relativity, you can go straight to Section 1.2. The present section can be skipped on a first reading.

1.1.1 Galilean transformations, Maxwell’s equations

Imagine standing on the ground and watching a train travel by with constant speed v in the x direction. Let the reference frame of the ground be labeled S, and let the reference frame of the train be labeled S

^′

, as shown in Fig. 1.1. Consider two events that happen v

x y

(ground) (train) z

x' y'

S z' S'

Figure

1.1

on the train. An event is defined as something that occurs with definite space and time coordinates (as measured in a given frame). For example, a person might clap her hands; this clap takes place at a definite time and a definite location. Technically, the clap lasts for a nonzero time (a few hundredths of a second), and the hands extend over a nonzero distance (a few inches). But we’ll ignore these issues and assume that the clap can be described be unique x, y, z, and t values. Note that a given event isn’t associated with one particular frame. The event simply happens, independent of a frame. For any arbitrary frame we then choose to consider, we can describe the event by specifying the coordinates as measured in that frame.

On our train, the two events might be one person clapping her hands and another person stomping his feet. If the space and time separations between these two events in the frame of the train are ∆x

^′

and ∆t

^′

, what are the space and time separations, ∆x and ∆t, in the frame of the ground? Ignoring what we’ll be learning about relativity in this chapter, the answers are “obvious” (although, as we’ll see in Section 2.1 when we derive the Lorentz transformations, obvious things can apparently be incorrect!). The time separation ∆t is “clearly” the same as on the train, so we have ∆t = ∆t

^′

. We know from everyday experience that nothing strange happens with time. When you see people exiting a train station, they’re not fiddling with their watches, trying to recalibrate them with a ground-based clock.

The spatial separation is a little more interesting, but still fairly simple. If the train

weren’t moving, then we would just have ∆x = ∆x

^′

. This is true because if the train

isn’t moving, then the only possible difference between the frames is the location of

the origin. But the only consequence of this difference is that every x

^′

coordinate in

the train is equal to a given fixed number plus the corresponding x coordinate on the

ground. This fixed number then cancels when calculating the separation, ∆x

^′

≡ x

₂^′

− x

^′₁

.

However, in the general case where the train is moving, everything in the train gets

carried along at speed v during the time ∆t (which equals ∆t

^′

) between the two events.

So as seen in the ground frame, the person stomping his feet ends up v ∆t

^′

to the right (or left, if v is negative) of where he would be if the train weren’t moving. The total spatial separation ∆x between the events in the ground frame is therefore the ∆x

^′

separation that would arise if the train weren’t moving, plus the v ∆t

^′

separation due to the motion of the train. That is, ∆x = ∆x

^′

+ v ∆t

^′

, as shown in Fig. 1.2.

v

v Event 1

(train end) (train start)

x' x'

z' z'

y' y'

v ∆t' ∆x'

∆x S' S'

x y (ground)

z S

Event 2

Figure

1.2

The Galilean transformations (first written down by Galileo Galilei in 1638) are therefore

∆x = ∆x

^′

+ v ∆t

^′

∆t = ∆t

^′

(1.1)

Nothing interesting happens in the y and z directions (assuming the train is traveling in the x direction), so we additionally have ∆y = ∆y

^′

and ∆z = ∆z

^′

. Both of these common values are zero for the events in Fig. 1.2, because the events occur on the x axis. But for general locations of the events, the ∆’s will be nonzero.

A special case of Eq. (1.1) arises when the two events occur at the same place on the train, so that ∆x

^′

= 0. In this case we have ∆x = v ∆t

^′

. This makes sense, because the spot on the train where the events occur simply travels a distance v ∆t (which equals v ∆t

^′

) by the time the second event occurs.

The principle of Galilean invariance says that the laws of physics are invariant under the above Galilean transformations. Alternatively, it says that the laws of physics hold in all inertial (non-accelerating) frames. (It was assumed prior to Einstein that these two statements say the same thing, but we will soon see that they do not. The second statement is the one that remains valid in relativity.) This principle is quite believable.

For example, in Galilean (nonrelativistic) physics, Newton’s second law, F = ma (or really F = dp/dt) holds in all inertial frames, because (1) the force F is the same in all inertial frames, and (2) the constant relative velocity v

_rel

between any two inertial frames implies that the acceleration of a given particle is the same in all inertial frames. Written out explicitly, the velocities v

₁

and v

₂

in the two frames are related by v

₁

= v

2

+ v

rel

, so

a

₁

≡ dv

₁

dt

₁

= d (v

₂

+ v

rel

) dt

₁

= dv

₂

dt

₂

+ 0 ≡ a

2

, (1.2)

where we have used the facts that t

₁

= t

2

(at least in a Galilean world) and that the derivative of a constant is zero.

Remarks:

1. Note that the Galilean transformations in Eq. (1.1) aren’t symmetric in x and t. This isn’t

necessarily a bad thing, but it turns out that it will in fact be a problem in special relativity,

where space and time are treated on a more equal footing. We’ll find in Section 2.1 that the

Galilean transformations are replaced by the Lorentz transformations, and the latter are in

fact symmetric in x and t (up to factors of the speed of light, c).

2. Eq. (1.1) deals only with the differences in the x and t values between two events, and not with the values of the coordinates themselves of each event. The values of the coordinates of a single event depend on where you pick your origin, which is an arbitrary choice. The coordinate differences between two events, however, are independent of this choice, and this allows us to make the physically meaningful statements in Eq. (1.1). Since it makes no sense for a physical result to depend on your arbitrary choice of origin, the Lorentz transformations we derive in Section 2.1 will also need to involve only differences in coordinates.

3. We’ve been talking a lot about “events,” so just to make sure we’re on the same page with the definition of an event, we should give some examples of things that are not events. If a train is at rest on the ground (or even if it is moving), and if you look at it at a snapshot in time, then this doesn’t describe an event, because the train has spatial extent. There isn’t a unique spatial coordinate that describes the train. If you instead consider a specific point on the train at the given instant, then that does describe an event. As another example of a non event, if you look at a pebble on the ground for a minute, then this doesn’t describe an event, because you haven’t specified the time coordinate. If you instead consider the pebble at a particular instant in time, then that does describe an event. (We’ll consider the pebble to be a point object, so that the spatial coordinate is unique.) ♣

We introduced the Galilean transformations above because of their relation (more precisely, their conflict) with Maxwell’s equations. One of the great triumphs of 19th- century physics was the theory of electromagnetism. In 1864, James Clerk Maxwell wrote down a set of equations that collectively described everything that was known about the subject. These equations involve the electric and magnetic fields through their space and time derivatives. Maxwell’s original formulation consisted of a large number of equations, but these were later written more compactly, using vector calculus, as four equations. We won’t worry about their specific form here, but it turns out that if you transform the equations from one reference frame to another via the Galilean transformations, they end up taking a different form. That is, if you’ve written down Maxwell’s equations in one frame (where they take their standard nice-looking form), and if you then replace the coordinates in this frame by those in another frame, using Eq. (1.1), then the equations look different (and not so nice).

This different appearance presents a major problem. If Maxwell’s equations take a nice form in one frame and a not-so-nice form in every other frame, then why is one frame special? Said in another way, it can be shown that Maxwell’s equations imply that light moves with a certain speed c. But which frame is this speed measured with respect to? The Galilean transformations imply that if the speed is c with respect to a given frame, then it is not c with respect to any other frame. (You need to add or subtract the relative speed v between the frames.) The proposed special frame where Maxwell’s equations are nice and the speed of light is c was called the frame of the ether. We’ll talk in detail about the ether in the next subsection, but experiments showed that light was surprisingly always measured to move with speed c in every frame, no matter which way the frame was moving through the supposed ether. We say “supposed” because the final conclusion was that the ether simply doesn’t exist.

There were thus various possibilities. Something was wrong with either Maxwell’s

equations, the Galilean transformations, or the way in which measurements of speed

were done (see Footnote 2 on page 8). Considering how “obvious” the Galilean trans-

formations are, the natural assumption in the late 19th century was that the problem lay

elsewhere. However, after a good deal of effort by many people to make everything

else fit with the Galilean transformations, Einstein finally showed that these were in

fact the culprit. It was well known that Maxwell’s equations were invariant under the

Lorentz transformations (in contrast with their non-invariance under the Galilean ones),

but Einstein was the first to recognize the full meaning of these transformations. Instead of being relevant only to electromagnetism, the Lorentz transformations replaced the Galilean ones universally.

More precisely, in 1905 Einstein showed why the Galilean transformations are simply a special case of the Lorentz transformations, valid (to a high degree of accuracy) only when the speed involved is much less than the speed of light. As we’ll see in Section 2.1, the coefficients in the Lorentz transformations depend on both the relative speed v of the frames and the speed of light c, where the c’s appear in various denominators. Since c is quite large (about 3 · 10

⁸

m /s) compared with everyday speeds v, the parts of the Lorentz transformations involving c are negligible, for any typical v. The surviving terms are the ones in the Galilean transformations in Eq. (1.1). These are the only terms that are noticeable for everyday speeds. This is why no one prior to Einstein realized that the correct transformations between two frames had anything to do with the speed of light.

As he pondered the long futile fight To make Galileo’s world right, In a new variation

Of the old transformation,

It was Einstein who first saw the light.

In short, the reasons why Maxwell’s equations are in conflict with the Galilean transformations are: (1) The speed of light is what determines the scale at which the Galilean transformations break down, (2) Maxwell’s equations inherently involve the speed of light, because light is an electromagnetic wave.

1.1.2 Michelson–Morley experiment

As mentioned above, it was known in the late 19th century, after Maxwell wrote down his equations, that light is an electromagnetic wave and that it moves with a speed of about 3 · 10

⁸

m /s.1 Now, every other wave that people knew about at the time needed a medium to propagate in. Sound waves need air, ocean waves of course need water, waves on a string of course need the string, and so on. It was therefore natural to assume that light also needed a medium to propagate in. This proposed medium was called the ether.

However, if light propagates in a given medium, and if the speed in this medium is c, then the speed in a reference frame moving relative to the medium should be different from c. Consider, for example, sound waves in air. If the speed of sound in air is v

_sound

, and if you run toward a sound source with speed v

_you

, then the speed of the sound waves with respect to you (assuming it’s a windless day) is v

_sound

+ v

you

. Equivalently, if you are standing at rest downwind and the speed of the wind is v

_wind

, then the speed of the sound waves with respect to you is v

_sound

+ v

wind

.

Assuming that the ether really exists (although we’ll soon see that it doesn’t), a reasonable thing to do is to try to measure one’s speed with respect to it. This can be done as follows. We’ll frame this discussion in terms of sound waves in air. Let v

_s

be the speed of sound in air. Imagine two people standing on the ends of a long platform of length L that moves at speed v

_p

with respect to the reference frame in which the air is at rest. One person claps, the other person claps immediately when he hears the first clap (assume that the reaction time is negligible), and then the first person records the

1The exact value of the speed is 299,792,458 m/s. A meter is actually defined to be 1/299,792,458 of the distance that light travels in one second in vacuum. So this speed of light is exact. There is no need for an error bar because there is no measurement uncertainty.

total time elapsed when she hears the second clap. What is this total time? Well, we can’t actually give an answer without knowing which direction the platform is moving.

Is it moving parallel to its length, or perpendicular to it (or somewhere in between)?

Let’s look at these two basic cases. For both of these, we’ll view the setup and do the calculation in the frame in which the air is at rest.

• Parallel motion: Consider first the case where the platform moves parallel to its length. In the reference frame of the air, assume that the person at the rear is the one who claps first; see Fig. 1.3. Then if v

_s

is the speed of sound and v

_p

is the v

L v

_s p

Figure

1.3 speed of the platform, it takes a time of L /(v

s

− v

p

) for the sound from the first clap to travel forward to the front person. This is true because the sound closes the initial gap of L at a relative speed of v

_s

− v

p

, as viewed in the frame of the air. (Alternatively, relative to the initial position of the back of the platform, the position of the sound wave is v

_s

t, and the position of the front person is L + v

p

t.

Equating these gives t = L/(v

s

−v

p

).) This time is longer than the naive answer of L /v

s

because the front person is moving away from the rear person, which means that the sound has to travel farther than L.

By similar reasoning, the time for the sound from the second clap to travel backward to the rear person is L /(v

s

+ v

p

). This time is shorter than the naive answer of L /v

s

because the rear person is moving toward the front person, which means that the sound travels less than L.

Adding the forward and backward times gives a total time of t

₁

= L

v

_s

− v

p

+ L

v

_s

+ v

p

= 2Lv

_s

v

_s²

− v

²p

. (1.3)

This correctly equals 2L /v

s

when v

_p

= 0. In this case the platform is at rest, so the sound simply needs to travel forward and backward a total distance of 2L at speed v

_p

. And the result correctly equals infinity when v

_p

→ v

s

. In this case the front person is receding as fast as the sound is traveling, so the sound from the first clap can never catch up.

• Perpendicular motion: Now consider the case where the platform moves perpendicular to its length. In the reference frame of the air, we have the situation shown in Fig. 1.4. The sound moves diagonally with speed v

_s

. (The sound actually

v

_p

v

_p

v

L

p

v

_s

v

_s

platform

Figure

1.4

moves in all directions, of course, but it’s only the part of the sound wave that moves in a particular diagonal direction that ends up hitting the other person.) Since the “horizontal” component of the diagonal velocity is the platform’s speed v

_p

, the Pythagorean theorem gives the “vertical” component as √

v

_s²

− v

p2

, as shown in Fig. 1.5. This is the speed at which the length L of the platform is

v

_p

v

_s

v

_p²

v

_s²

-

Figure

1.5

traversed during both the out and back parts of the trip. So the total time is t

₂

= 2L

√ v

²_s

− v

p²

. (1.4)

Again, this correctly equals 2L /v

s

when v

_p

= 0, and infinity when v

p

→ v

s

. The vertical component of the velocity is zero in the latter case, because the diagonal path is essentially horizontal.

The times in Eqs. (1.3) and (1.4) are not equal; you can quickly show that t

₁

≥ t

2

.

It turns out that (for given values of v

_s

and v

_p

) of all the possible orientations of the

platform relative to the direction of motion (which we have been taking to be rightward),

the t

₁

in Eq. (1.3) is the largest possible time, and the t

₂

in Eq. (1.4) is the smallest.

(The proof of this is somewhat tedious, but at least it is believable that if the platform is oriented between the above two special cases, the time lies between the associated times t

₁

and t

₂

.) Therefore, if you are on a large surface that is moving with respect to the air, and if you know the value of v

_s

, then if you want to figure out what v

_p

is, all you have to do is repeat the above experiment with someone standing at various points along the circumference of a given circle of radius L around you. (Assume that it doesn’t occur to you to toss a little piece of paper in the air, in order to at least determine the direction of the wind with respect to you.) If you take the largest total time observed and equate it with t

₁

, then Eq. (1.3) will give you v

_p

. Alternatively, you can equate the smallest total time with t

₂

, and Eq. (1.4) will yield the same v

_p

.

In the limiting case where v

_p

≪ v

s

, we can make some approximations to the above expressions for t

₁

and t

₂

. These approximations involve the Taylor-series expressions 1 /(1 − ϵ) ≈ 1 + ϵ and 1/ √

1 − ϵ ≈ 1 + ϵ/2. (See Appendix G for a discussion of Taylor series.) These expressions yield the following approximate result for the difference between t

₁

and t

₂

(after first rewriting t

₁

and t

₂

so that a “1” appears in the denominator):

∆t = t

1

− t

2

= 2L v

_s

*. .

, 1 1 − v

p²

/v

s²

− 1

√

1 − v

p²

/v

s²

+/ / -

≈ 2L v

_s

*

, (

1 + v

_p²

v

_s²

)

− (

1 + v

_p²

2v

_s²

) + -

= Lv

_p²

v

_s³

. (1.5)

The difference t

₁

− t

2

is what we’ll be concerned with in the Michelson–Morley experi- ment, which we will now discuss.

The strategy in the above sound-in-air setup is the basic idea behind Michelson’s and Morley’s attempt in 1887 to measure the speed of the earth through the supposed ether. (See Handschy (1982) for the data and analysis of the experiment.) There is, however, a major complication with light that doesn’t arise with sound. The speed of light is so large that any time intervals that are individually measured will inevitably have measurement errors that are far larger than the difference between t

₁

and t

₂

. Therefore, individual time measurements give essentially no information. Fortunately, there is a way out of this impasse. The trick is to measure t

₁

and t

₂

concurrently, as opposed to separately. More precisely, the trick is to measure only the difference t

₁

− t

2

, and not the individual values t

₁

and t

₂

. This can be done as follows.

Consider two of the above “platform” scenarios arranged at right angles with re- spect to each other, with the same starting point. This can be arranged by having a (monochromatic) light beam encounter a beam splitter that sends two beams off at 90

^◦

angles. The beams then hit mirrors and bounce back to the beam splitter where they (partially) recombine before hitting a screen; see Fig. 1.6. The fact that light is a wave,

mirror

mirror beam

splitter source

screen

recombined beam

Figure

1.6 which is what got us into this ether mess in the first place, is now what saves the day. The

wave nature of light implies that the recombined light beam produces an interference pattern on the screen. At the center of the pattern, the beams will constructively or destructively interfere (or something in between), depending on whether the two light beams are in phase or out of phase when they recombine. This interference pattern is extremely delicate. The slightest change in travel times of the beams will cause the pattern to noticeably shift. This type of device, which measures the interference between two light beams, is known as an interferometer.

If the whole apparatus is rotated around, so that the experiment is performed at

various angles, then the maximum amount that the interference pattern changes can be

used to determine the speed of the earth through the ether (v

_p

in the platform setup above). In one extreme case, the time in a given arm is longer than the time in the other arm by Lv

²

/c

³

. (We have changed notation in Eq. (1.5) so that v

_p

→ v is the speed of the earth through the supposed ether, and v

_s

→ c is the speed of light.) But in the other extreme case, the time in the given arm is shorter by Lv

²

/c

³

. So the maximum shift in the interference pattern corresponds to a time difference of 2Lv

²

/c

³

.

However, when Michelson and Morley performed their experiment, they observed no interference shift as the apparatus was rotated around. Their setup did in fact allow enough precision to measure a nontrivial earth speed through the ether, if such a speed existed. So if the ether did exist, their results implied that the speed of the earth through it was zero. This result, although improbable, was technically fine. It might simply have been the case that they happened to do their experiment when the relative speed was zero. However, when they performed their experiment a few months later, when the earth’s motion around the sun caused it to be moving in a different direction, they still measured zero speed. It wasn’t possible for both of these results to be zero (assuming that the ether exists), without some kind of modification to the physics known at the time.

Many people over the years tried to explain this null result, but none of the expla- nations were satisfactory. Some led to incorrect predictions in other setups, and some seemed to work fine but were a bit ad hoc.2 The correct explanation, which followed from Einstein’s 1905 theory of relativity, was that the ether simply doesn’t exist.3 In other words, light doesn’t need a medium to propagate in. It doesn’t move with respect to a certain special reference frame, but rather it moves with respect to whoever is looking at it.

The findings of Michelson–Morley Allow us to say very surely,

“If this ether is real, Then it has no appeal,

And shows itself off rather poorly.”

Remarks:

1. We assumed above that the lengths of the two arms in the apparatus were equal. However, in practice there is no hope of constructing lengths that are equal, up to errors that are small compared with the wavelength of the light. But fortunately this doesn’t matter. We’re concerned not with the difference in the travel times associated with the two arms, but rather with the difference in these differences as the apparatus is rotated around. Using Eqs. (1.3) and (1.4) with different lengths L

₁

and L

₂

, you can show (assuming v ≪ c) that the maximum interference shift corresponds to a time of (L

₁

+ L

2

)v

²

/c

³

. This is the generalization of the 2Lv

²

/c

³

result we derived in Eq. (1.5) (in different notation) when the lengths were equal. The measurement errors in L

₁

and L

₂

therefore need only be small compared with the (macroscopic) lengths L

₁

and L

₂

, as opposed to small compared with the (microscopic) wavelength of light.

2. Assuming that the lengths of the arms are approximately equal, let’s plug in some rough numbers to see how much the interference pattern shifts. The Michelson–Morley setup

2The most successful explanation was the Lorentz–FitzGerald contraction. These two physicists indepen- dently proposed that lengths are contracted in the direction of the motion by precisely the right factor, namely

√1− v²/c², to make the travel times in the two arms of the Michelson–Morley setup equal, thus yielding the null result. This explanation was essentially correct, although the reason why it was correct wasn’t known until Einstein came along.

3Although we’ve presented the Michelson–Morley experiment here for pedagogical purposes, the consen- sus among historians is that Einstein actually wasn’t influenced much by the experiment, except indirectly through Lorentz’s work on electrodynamics. See Holton (1988).

had arms with effective lengths of about 10 m. We’ll take v to be on the order of the speed of the earth around the sun, which is about 3 · 10

⁴

m /s. We then obtain a maximal time difference of t = 2Lv

²

/c

³

≈ 7 · 10

⁻¹⁶

s. The large negative exponent here might make us want to throw in the towel, thinking that the effect is hopelessly small. However, the distance that light travels in the time t is ct = (3 · 10

⁸

m/s)(7 · 10

⁻¹⁶

s) ≈ 2 · 10

⁻⁷

m, and this happens to be a perfectly reasonable fraction of the wavelength of visible light, which is around λ = 6 · 10

⁻⁷

m, give or take. So we have ct/λ ≈ 1/3. This maximal interference shift of about a third of a cycle was well within the precision of the Michelson–Morley setup. So if the ether had really existed, Michelson and Morley definitely would have been able to measure the speed of the earth through it.

3. One proposed explanation of the observed null effect was “frame dragging.” What if the earth drags the ether along with it, thereby always yielding the observed zero relative speed? This frame dragging is quite plausible, because in the platform example above, the platform drags a thin layer of air along with it. And more mundanely, a car completely drags the air in its interior along with it. But it turns out that frame dragging is inconsistent with stellar aberration, which is the following effect.

Depending on the direction of the earth’s instantaneous velocity as it obits around the sun, it is an experimental fact that a given star might (depending on its location) appear at slightly different places in the sky when viewed at two times, say, six months apart. This is due to the fact that a telescope must be aimed at a slight angle relative to the actual direction to the star, because as the star’s light travels down the telescope, the telescope moves slightly in the direction of the earth’s motion. We’re assuming (correctly) here that frame dragging does not exist.

As a concrete analogy, imagine holding a tube while running through vertically falling rain. If you hold the tube vertically, then the raindrops that enter the tube won’t fall cleanly through. Instead, they will hit the side of the tube, because the tube is moving sideways while the raindrops are falling vertically. However, if you tilt the tube at just the right angle, the raindrops will fall (vertically) cleanly through without hitting the side. At what angle θ should the tube be tilted? If the tube travels horizontally a distance d during the time it takes a raindrop to fall vertically a distance h, then the ratio of these distances must equal the ratio of the speeds: d /h = v

tube

/v

rain

(see Fig. 1.7). The angle θ is then given

v d

h

tube

(start) (end)

θ

v

_rain

Figure

1.7 by tan θ = d/h =⇒ tan θ = v

tube

/v

rain

. With respect to your frame as you run along, the

raindrops come down at an angle θ; they don’t come down vertically.

Returning to the case of light, v

_tube

gets replaced with (roughly) the speed v of the earth around the sun, and v

_rain

gets replaced with the speed c of light. The ratio of these two speeds is about v /c = 10

⁻⁴

, so the effect is small. But it is large enough to be noticeable, and it has indeed been measured; stellar aberration does exist. At two different times of the year, a telescope must be pointed at slightly different angles when viewing a given star.

Now, if frame dragging did exist, then the light from the star would get dragged along with the earth and would therefore travel down a telescope that was pointed directly at the star, in disagreement with the observed fact that the telescope must point at the slight angle mentioned above. (Or even worse, the dragging might produce a boundary layer of turbulence that would blur the stars.) The existence of stellar aberration therefore implies that frame dragging doesn’t occur.

4. Note that it is the velocity of the telescope that matters in stellar aberration, and not its

position. This aberration effect should not be confused with the parallax effect, where

the direction of the actual position of an object changes, depending on the position of the

observer. For example, people at different locations on the earth see the moon at slightly

different angles (that is, they see the moon in line with different distant stars). As a more

down-to-earth example, two students sitting at different locations in a classroom see the

teacher at different angles. Although stellar parallax has been measured for nearby stars

(as the earth goes around the sun), its angular effect is much smaller than the angular effect

from stellar aberration. The former decreases with the distance to the star, whereas the

latter doesn’t. For further discussion of aberration, and of why it is only the earth’s velocity

(or rather, the change in its velocity) that matters, and not also the star’s velocity (since

you might think, based on the fact that we’re studying relativity here, that it is the relative velocity that matters), see Eisner (1966). ♣

1.2 The postulates

Let’s now start from scratch and see what the theory of special relativity is all about.

We’ll take the route that Einstein took and use two postulates as the foundation of the theory. We’ll start with the “relativity postulate” (also called the Principle of Relativity). This postulate is quite believable, so you might just take it for granted and forget to consider it. But like any other postulate, it is crucial. It can be stated in various ways, but we’ll write it simply as:

• Postulate 1: All inertial (non-accelerating) frames are “equivalent.”

This postulate says that a given inertial frame is no better than any other; there is no preferred reference frame. That is, it makes no sense to say that something is moving. It makes sense only to say that one thing is moving with respect to another. This is where the “relativity” in “special relativity” comes from. There is no absolute inertial frame;

the motion of any frame is defined only relative to other frames.

This postulate also says that if the laws of physics hold in one inertial frame (and presumably they do hold in the frame in which I now sit), then they hold in all others.

(Technically, the earth is spinning while revolving around the sun, and there are also little vibrations in the floor beneath my chair, etc., so I’m not really in an inertial frame.

But it’s close enough for me.) The postulate also says that if we have two frames S and S

^′

, then S should see things in S

^′

in exactly the same way as S

^′

sees things in S, because we can just switch the labels of S and S

^′

. (We’ll get our money’s worth out of this statement in the next few sections.) It also says that empty space is homogeneous (that is, all points look the same), because we can pick any point to be, say, the origin of a coordinate system. It also says that empty space is isotropic (that is, all directions look the same), because we can pick any axis to be, say, the x axis of a coordinate system.

Unlike the second postulate below (the speed-of-light postulate), this first one is entirely reasonable. We’ve gotten used to having no special places in the universe. We gave up having the earth as the center, so let’s not give any other point a chance, either.

Copernicus gave his reply

To those who had pledged to deny.

“All your addictions To ancient convictions

Won’t bring back your place in the sky.”

The first postulate is nothing more than the familiar principle of Galilean invariance, assuming that this principle is written in the “The laws of physics hold in all inertial frames” form, and not in the form that explicitly mentions the Galilean transformations in Eq. (1.1), which are inconsistent with the second postulate below.

Everything we’ve said here about the first postulate refers to empty space. If we have a chunk of mass, then there is certainly a difference between the position of the mass and a point a meter away. To incorporate mass into the theory, we would have to delve into the subject of general relativity. But we won’t have anything to say about that in this chapter. We will deal only with empty space, containing perhaps a few observant souls sailing along in rockets or floating aimlessly on little spheres. Although that might sound boring at first, it will turn out to be anything but.

The second postulate of special relativity is the “speed-of-light” postulate. This one

is much less intuitive than the relativity postulate.

• Postulate 2: The speed of light in vacuum has the same value c (approximately 3 · 10

⁸

m /s) in any inertial frame.

This statement certainly isn’t obvious, or even believable. But on the bright side, at least it’s easy to understand what the postulate says, even if you think it’s too silly to be true.

It says the following. Consider a train moving along the ground with constant velocity.

Someone on the train shines a light from one point on the train to another. The speed of the light with respect to the train is c. Then the above postulate says that a person on the ground also sees the light move at speed c.

This is a rather bizarre statement. It doesn’t hold for everyday objects. If a baseball is thrown forward with a given speed on a train, then the speed of the baseball is different in the ground frame. An observer on the ground must add the velocity of the ball (with respect to the train) to the velocity of the train (with respect to the ground) in order to obtain the velocity of the ball with respect to the ground. Strictly speaking, this isn’t quite true, as the velocity-addition formula in Section 1.5 shows. But it’s true enough for the point we’re making here.

The truth of the speed-of-light postulate cannot be demonstrated from first princi- ples. No statement with any physical content in physics (that is, one that isn’t purely mathematical, such as, “two apples plus two apples gives four apples”) can be proven.

In the end, we must rely on experiment. And indeed, all of the consequences of the speed-of-light postulate have been verified countless times during the past century. As discussed in the previous section, the most well-known of the early experiments on the speed of light was the one performed by Michelson and Morley. The zero interference shift they always observed implied that the v

_p

speed in Eq. (1.5) was always zero. This in turn implies that no matter what (inertial) frame you are in, you are always at rest with respect to a frame in which the speed of light is c. In other words, the speed of light is the same in any inertial frame.

In more recent years, the consequences of the second postulate have been verified continually in high-energy particle accelerators, where elementary particles reach speeds very close to c. The collection of all the data from numerous experiments over the years allows us to conclude with near certainty that our starting assumption of an invariant speed of light is correct (or is at least the limiting case of a more correct theory).

Remark: Given the first postulate, you might wonder if we even need the second. If all inertial frames are equivalent, shouldn’t the speed of light be the same in any frame? Well, no. For all we know, light might behave like a baseball. A baseball certainly doesn’t have the same speed in all inertial frames, and this doesn’t ruin the equivalence of the frames.

It turns out (see Section 2.7) that nearly all of special relativity can be derived by invoking

only the first postulate. The second postulate simply fills in the last bit of necessary information

by stating that something has the same finite speed in every frame. It’s actually not important that this thing is light. It could be mashed potatoes or something else, and the theory would still come out the same. (Well, the thing has to be massless, as we’ll see in Chapter 3, so we’d need to have massless potatoes, but whatever.) The second postulate can therefore be stated more minimalistically as, “There is something that has the same speed in any inertial frame.” It just so happens that in our universe this thing is what allows us to see.

To go a step further, it’s not even necessary for there to exist something that has the same speed in any frame. The theory will still come out the same if we write the second postulate as,

“There is a limiting speed of an object in any frame.” (See Section 2.7 for a discussion of this.) There is no need to have something that actually travels at this speed. It’s conceivable to have a theory that contains no massless objects, in which case everything travels slower than the limiting speed. ♣

Let’s now see what we can deduce from the above two postulates. There are many

different ways to arrive at the various kinematical consequences. Our road map for

the initial part of the journey (through Section 2.2) is shown in Fig. 1.8. Additional kinematics topics are covered in Sections 2.3 through 2.7.

Section 2.1.1

Section 2.1.3

Section 1.3

Section 1.5 Section 2.2

Appendix D

The two postulates Velocity addition

Lorentz transformations Fundamental effects

Figure

1.8

1.3 The fundamental effects

The most striking effects of the two postulates of relativity are (1) the loss of simultaneity (equivalently, the rear-clock-ahead effect), (2) time dilation, and (3) length contraction.

In this section, we’ll discuss these three effects by using some time-honored concrete setups. In Chapter 2, we’ll use these three effects to derive the Lorentz transformations.

1.3.1 Loss of simultaneity

The basic effect

Consider the following setup. In person A’s reference frame, a light source is placed midway between two receivers, a distance ℓ from each (see Fig. 1.9). The light source

c c

l l

A

v B

Figure

1.9

emits a flash. In A’s reference frame, the light hits the two receivers at the same time, ℓ/c seconds after the flash. So if Event 1 is “light hitting the left receiver” and Event 2 is “light hitting the right receiver,” then the two events are simultaneous in A’s frame.

Now consider another observer, B, who travels to the left at speed v. In B’s reference frame, does the light hit the receivers at the same time? That is, are Events 1 and 2 simultaneous in B’s frame? We will show that surprisingly they are not.

In B’s reference frame, the situation looks like that in Fig. 1.10. If you want, you

c c

v v

B

A l' l'

Figure

1.10

can think of A as being on a train, and B as standing on the ground. With respect to B, the receivers (along with everything else in A’s frame) move to the right with speed v. Additionally, with respect to B (and this is where the strangeness of relativity comes into play), the light travels in both directions at speed c, as indicated in the figure. Why is this the case? Because the speed-of-light postulate says so!

Note that everyday objects do not behave this way. Consider, for example, a train

(A’s frame) moving at 30 mph with respect to the ground (B’s frame). If A stands in

the middle of the train and throws two balls forward and backward, each with speed 50

mph with respect to the train, then the speeds of the two balls with respect to the ground

are 50 − 30 = 20 mph (backward) and 50 + 30 = 80 mph (forward). (We’re ignoring

the minuscule corrections from the velocity-addition formula discussed in Section 1.5.)

These two speeds are different. In contrast with everyday objects like these balls, light has the bizarre property that its speed is always c (when viewed in an arbitrary inertial frame), independent of the speed of the source. Strange, but true.

Returning to our setup with the light beams and receivers, we can say that because B sees both light beams move with speed c, the relative speed (as viewed by B) of the light and the left receiver is c + v, and the relative speed (as viewed by B) of the light and the right receiver is c − v.

Remark: Yes, it is legal to simply add or subtract these speeds to obtain the relative speeds as

viewed by B. The reasoning here is the same as in the discussion of Fig. 1.3 in Section 1.1.2, where

we obtained relative speeds of v

s

± v

p

. As a concrete example, if the v here equals 2 · 10

⁸

m/s, then in one second the left receiver moves 2 · 10

⁸

m to the right, while the left ray of light moves 3 · 10

⁸

m to the left. This means that they are now 5 · 10

⁸

m closer than they were a second ago.

In other words, the relative speed (as measured by B) is 5 · 10

⁸

m /s, which is c + v here. This is the rate at which the gap between the light and the left receiver closes. So in addition to calling it the “relative” speed, you can also call it the “gap-closing” speed. Note that the above reasoning implies that it is perfectly legal for the relative speed of two things, as measured by a third, to take any value up to 2c.

Likewise, the relative speed between the light and the right receiver is c − v. The v and c in these results are measured with respect to the same person, namely B, so our intuition involving simple addition and subtraction works fine. Even though we’re dealing with a relativistic speed v here, we don’t need to use the velocity-addition formula from Section 1.5, which is relevant in a different setting. This remark is included just in case you’ve seen the velocity-addition formula and think it’s relevant in this setup. But if it didn’t occur to you, then never mind.

Note that the speed of the right photon4 is not c −v. (And likewise the speed of the left photon is not c + v.) The photon moves at speed c, as always. It is the relative speed (as measured by B) of the photon and the front of the train that is c − v. No thing is actually moving with this speed in our setup. This speed is just the rate at which the gap closes. And a gap isn’t an actual moving thing. ♣

Let ℓ

^′

be the distance from the light source to each of the receivers, as measured by B.5 Then in B’s frame, the gap between the light beam and the left receiver starts with length ℓ

^′

and subsequently decreases at a rate c + v. The time for the light to hit the left receiver is therefore ℓ

^′

/(c + v). Similar reasoning holds for the right receiver along with the relative speed of c − v. The times t

L

and t

_R

at which the light hits the left and right receivers are therefore given by

t

_L

= ℓ

^′

c + v and t

_R

= ℓ

^′

c − v . (1.6)

These two times are not equal if v , 0. (The one exception is when ℓ

^′

= 0, in which case the two events happen at the same place and same time in all frames.) Since t

_L

< t

R

, we see that in B’s frame, the light hits the left receiver before it hits the right receiver. We have therefore arrived at the desired conclusion that the two events (light hitting back, and light hitting front) are not simultaneous in B’s frame.

The moral of this is that it makes no sense to say that one event happens at the same time as another, unless you also state which frame you’re dealing with. Simultaneity depends on the frame in which the observations are made.

4Photons are what light is made of. So “speed of the photon” means the same thing as “speed of the light beam.” Sometimes it’s easier to talk in terms of photons.

5We’ll see in Section 1.3.3 thatℓ^′is not equal to theℓ in A’s frame, due to length contraction. But this won’t be important for what we’re doing here. The only fact we need for now is that the light source is equidistant from the receivers, as measured by B. This is true because space is homogeneous, which implies that any length-contraction factor we eventually arrive at must be the same everywhere. More on this in Section 1.3.3.

Of the many effects, miscellaneous, The loss of events, simultaneous, Allows B to claim

There’s no pause in A’s frame, Remarks:

1. The strangeness of the t

_L

< t

R

loss-of-simultaneity result can be traced to the strangeness of the speed-of-light postulate. We entered the bizarre world of relativity when we wrote the c’s above the photons in Fig. 1.10. The t

_L

< t

R

result is a direct consequence of the nonintuitive fact that light moves with the same speed c in every inertial frame.

2. The invariance of the speed of light led us to the fact that the relative speeds between the photons and the left and right receivers are c + v and c − v. If we were talking about baseballs instead of light beams, then the relative speeds wouldn’t take these general forms.

If v

_b

is the speed at which the baseballs are thrown in A’s frame, then as we noted above in the case where v = 30 mph and v

b

= 50 mph, B sees the balls move with speeds of (essentially, ignoring the tiny corrections due to the velocity-addition formula) v

_b

− v to the left (assuming v

_b

> v) and v

b

+ v to the right. These are not equal (in contrast with what happens with light). By the same “gap-closing” reasoning we used above, the relative speeds (as viewed by B) between the balls and the left and right receivers are then (v

_b

−v) +v = v

b

and (v

_b

+v) −v = v

b

. These are equal, so B sees the balls hit the receivers at the same time, as we know very well from everyday experience.

3. As explained in the remark prior to Eq. (1.6), it is indeed legal to obtain the times in Eq. (1.6) by simply dividing ℓ

^′

by the relative speeds, c + v and c − v. The gaps start with length ℓ

^′

and then decrease at these rates. But if you don’t trust this, you can use the following reasoning. In B’s frame, the position of the right photon (relative to the initial position of the light source) simply equals ct, and the position of the right receiver (which has a head start of ℓ

^′

) equals ℓ

^′

+ vt. The photon hits the receiver when these two positions are equal. Equating them gives

ct

= ℓ

^′

+ vt =⇒ t

R

= ℓ

^′

c

− v . (1.7)

Similar reasoning with the left photon gives t

_L

= ℓ

^′

/(c + v).

4. There is always a difference between the time that an event happens and the time that someone sees the event happen, because light takes time to travel from the event to the observer. What we calculated above were the times t

_L

and t

_R

at which the events actually

happen in B’s frame. (These times are independent of where B is standing at rest in the

frame.) If we wanted to, we could calculate the times at which B sees the events occur.

(These times do depend on where B is standing at rest in the frame.) But such times are rarely important, so in general we won’t be concerned with them. They can easily be calculated by adding on a (distance)/c time for the photons to travel to B’s eye. Of course, if B actually did the above experiment to find t

_L

and t

_R

, she would do it by writing down the times at which she sees the events occur, and then subtracting off the relevant (distance)/c times, to find when the events actually happened.

To sum up, the t

_L

, t

R

result in Eq. (1.6) is due to the fact that the events truly occur at different times in B’s frame. The t

_L

, t

R

result has nothing to do with the time it takes

light to travel to B’s eye.

♣

Where this last line is not so extraneous.

The “rear clock ahead” effect

We showed in Eq. (1.6) that t

_L

is not equal to t

_R

, that is, the light hits the receivers at

different times in B’s frame. Let’s now be quantitative and determine the degree to which

two events that are simultaneous in one frame are not simultaneous in another frame.

Given the times t

_L

and t

_R

that we found in Eq. (1.6), the simplest quantitative number that we can produce, as a measure of the non-simultaneity, is the difference t

_R

−t

L

. This tells us how unsimultaneous the events are in B’s frame (the ground frame), given that they are simultaneous in A’s frame (the train frame). The interpretation of the resulting expression for t

_R

−t

L

is the task of Problem 1.1 (which relies on time dilation and length contraction, discussed below). But let’s take a slightly different route here, which will end up being a little more useful. This route will lead us to the rear-clock-ahead effect, which is the standard quantitative statement of the loss of simultaneity.

Consider a setup where two clocks are positioned at the ends of a train of length L (as measured in its own frame). The clocks are synchronized in the train frame. That is, they have the same reading at any given instant, as observed in the train frame, as you would naturally expect. (Throughout this book, we will always assume that clocks are synchronized in the frame in which they are at rest.) The train travels past you at speed v. It turns out that if you observe the clocks at simultaneous times in your frame, the readings will not be the same. You will observe the rear clock showing a higher reading than the front clock, as indicated in Fig. 1.11.

v

L

Figure

1.11 We’ll explain why this is true momentarily, but first let us note that a nonzero

difference in the readings is certainly a manifestation of the loss of simultaneity. To see why, consider a given instant in your (ground) frame when the rear and front clocks read, say, 12:01 and 12:00. (We’ll find that the actual difference depends on L and v, but let’s just assume it’s one minute here, for concreteness.) Assume that you hit both clocks simultaneously (in your ground frame) with paintballs when they show these readings.

Then in the train frame, the front clock gets hit when it reads 12:00, and then a minute later the rear clock gets hit when it reads 12:01. The simultaneous hits in your frame are therefore not simultaneous in the train frame. We have used the fact that the reading on a clock when a paintball hits it is frame independent. This is true because you can imagine that a clock breaks when a ball hits it, so that it remains stuck at a certain value.

Everyone has to agree on what this value is.

Let’s now find the exact difference in the readings on the two train clocks in Fig. 1.11.

To do this, we will (as we did above in Fig. 1.9) put a light source on the train. But we’ll now position it so that the light hits the clocks at the ends of the train at the same time in your (ground) frame. As in the discussion of Fig. 1.10, the relative speeds of the photons and the clocks are c + v and c − v (as viewed in your frame). We therefore need to divide the train into lengths in this ratio, in your frame, if we want the light to hit the ends at the same time. Now, because length contraction (discussed below in Section 1.3.3) is independent of position, the ratio in the train frame must also be c + v to c − v. You can then quickly show that two numbers that are in this ratio, and that add up to L, are L(c + v)/2c and L(c − v)/2c. (Mathematically, you’re solving the system of equations, x /y = (c + v)/(c − v) and x + y = L.) Dividing the train into these two

lengths (in the train frame, as shown in Fig. 1.12) causes the light to hit the ends of the L (c+v) 2c

L (c-v) _____

_____

2c

Figure

1.12 train simultaneously in the ground frame.

Let’s now examine what happens during the process in the train frame. Compared with the forward-moving light, Fig. 1.12 tells us that the backward-moving light must travel an extra distance of L(c +v)/2c − L(c −v)/2c = Lv/c. The light travels at speed c (as always), so the extra time is Lv /c

²

. The rear clock therefore reads Lv /c

²

more when it is hit by the backward photon, compared with what the front clock reads when it is hit by the forward photon. (Remember that the clocks are synchronized in the train frame.) This difference in readings has a frame-independent value, because the readings on the clocks when the photons hit them are frame independent, by the same reasoning as with the paintballs above.

Finally, let’s switch back to your (ground) frame. Let the instant you look at the

clocks be the instant the photons hit them. (That’s why we constructed the setup with the hittings being simultaneous in your frame.) Then from the previous paragraph, we conclude that you observe the rear clock reading more than the front clock by an amount Lv /c

²

:

The rear clock is ahead by Lv /c

²

. (1.8) This result is important enough to spell out in full and put in a box:

Rear-clock-ahead: If a train with length L moves with speed v relative to you, then you observe the rear clock reading Lv /c

²

more than the front clock, at any given instant.

This statement corresponds to Fig. 1.13. For concreteness, we have chosen the front

Lv

0

c2

__

v

Figure

1.13

clock to read zero. But if the front clock reads, say, 9:47, then the rear clock reads 9:47 plus Lv /c

²

. There is of course no need to have an actual train in the setup. In general, all we need are two clocks separated by some distance L and moving with the same speed v. But we’ll often talk in terms of trains, since they’re easy to visualize.

Note that the L in the Lv /c

²

result is the length of the train in its own frame, and not the shortened length that you observe in your frame (see Section 1.3.3). Appendix B gives a number of other derivations of Eq. (1.8), although they rely on material we haven’t covered yet.

Example (Clapping first):

Two people stand a distance L apart along an east-west road.

They clap simultaneously in the ground frame. In the frame of a car driving eastward along the road, which person claps first?

Solution:

The eastern person claps first, for the following reason. Without loss of generality, let’s assume that clocks on the two people read zero when the claps happen.

Then a snapshot in the ground frame at the instant the claps happen is shown in Fig. 1.14.

v

L

0 0

(ground frame)

*clap* *clap*

car

Figure

1.14

We’ve drawn the car in the middle as it travels past, but its exact location is irrelevant.

Now consider a snapshot in the car frame at the instant the eastern (right) person claps.

This person’s clock reads zero when he claps, because that is a frame-independent fact.

Now, we can imagine that the two people are on a westward-traveling train, which means that the western person is the front person. By the rear-clock-ahead effect, the western (left) person’s clock is behind by Lv/c

²

. So it reads only −Lv/c

²

, as shown in Fig. 1.15.

-Lv

0

c2

__

v v

(car frame) car

*clap*

Figure

1.15

(As we will see many times throughout this chapter, drawing pictures is extremely helpful when solving relativity problems!) Since this clock hasn’t hit zero, the western person hasn’t clapped yet. The eastern person therefore claps first, as we claimed.

Remark: In the car frame, the distance between the two people is actually less than L (as we have indicated in Fig. 1.15), due to the length-contraction result we’ll derive in Section 1.3.3. But this doesn’t affect the result that the eastern person claps first. Similarly, the time-dilation result that we’ll derive in Section 1.3.2 is relevant if we want to determine exactly how long the car observer needs to wait for the western person to clap. (It will turn out to be longer than Lv/c

²

.) We’ll talk about these matters shortly. ♣

Remarks:

1. The Lv/c

²

result has nothing to do with the fact that the rear clock passes you at a later

time than the front clock passes you. The train could already be past you, or it could even

be moving directly toward or away from you. The rear clock will still be ahead by Lv/c

²

,

as observed in your frame.

2. The Lv/c

²

result does not say that you see the rear clock ticking at a faster rate than the front clock. They run at the same rate. (They both have the same time-dilation factor relative to you; see Section 1.3.2.) The rear clock is simply always a fixed time ahead of the front clock, as observed in your frame.

3. In the train setup (with the off-centered light source) that led to Eq. (1.8), the fact that the rear clock is ahead of the front clock in the ground frame means that in the train frame the light hits the rear clock after it hits the front clock.

4. The L in Eq. (1.8) is the separation between the clocks in the longitudinal direction, that is, the direction of the velocity of the train (or more generally, the velocity of the clocks, if we don’t have a train). The height in the train doesn’t matter; all clocks along a given line perpendicular to the train’s velocity have the same reading at any given instant in the ground frame.

5. For everyday speeds v, the Lv/c

²

effect is extremely small. If v = 30 m/s (about 67 mph) and if L = 100 m, then Lv/c

²

≈ 3 · 10

⁻¹⁴

s. This is completely negligible on an everyday scale.

6. What if we have a train that doesn’t contain the above setup with a light source and two light beams? That is, what if the given events have nothing to do with light? The Lv /c

²

result still holds, because we could have built the light setup if we wanted to (arranging for the light-hitting-end events to coincide with the given events). It doesn’t matter if the light setup actually exists.

7. It’s easy to forget which of the clocks is the one that is ahead. But a helpful mnemonic for remembering “rear clock ahead” is that both the first and fourth letters in each word form the same acronym, “rca,” which is an anagram for “car,” which is sort of like a train. Sure.

♣

1.3.2 Time dilation

We showed above that if two clocks are separated by a distance L in the horizontal (that is, longitudinal) direction on a train, and if the train is moving with respect to you, then you observe different readings on the clocks, at any given instant in your frame. Note that this result relates the readings on two different clocks at a given instant in your frame. It says nothing about the rate at which a single clock runs in your frame. This is what we will now address.

We will demonstrate that if a given clock is moving with respect to you, then you will observe the clock running slowly. That is, if you use a stopwatch to measure how long it takes a given train clock to tick off 10 seconds, your stopwatch might read 20 seconds. The exact time on your watch depends on the speed v of the train, as we’ll see shortly. But in any case, your clock will always read more than 10 seconds in this setup. (Or it will read exactly 10 seconds if v = 0 and the train is sitting at rest.) This effect is called time dilation. The name is appropriate, because the word “dilate” means to become larger. Since the moving clock runs slow (as viewed by you), a time T that ticks off on it takes more than a time T on your watch.

We will derive this time-dilation result by presenting a classic example of a light beam traveling in the vertical (that is, transverse) direction on a train. Let there be a light source on the floor of the train, and let there be a mirror on the ceiling, which is a height h above the floor. Let observer A be at rest on the train, and let observer B be at rest on the ground. The speed of the train with respect to the ground is v.6 A flash of light is emitted upward. The light travels up to the mirror, bounces off it, and then

6Technically, the words “with respect to . . . ” should always be included when talking about speeds, because there is no absolute reference frame, and hence no absolute speed. But in the future, when it is clear what we mean (as in the case of a train moving with respect to the ground), we’ll occasionally be sloppy and drop the “with respect to . . . .”